Chemistry
13th Edition
ISBN: 9781259911156
Author: Raymond Chang Dr., Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.160QP
One way to gain a physical understanding of b in the van der Waals equation is to calculate the “excluded volume.” Assume that the distance of closest approach between two similar atoms is the sum of their radii (2r). (a) Calculate the volume around each atom into which the center of another atom cannot penetrate. (b) From your result in (a), calculate the excluded volume for 1 mole of the atoms, which is the constant b. How does this volume compare with the sum of the volumes of 1 mole of the atoms?
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A sample of 3.54 mol of krypton is confined at low pressure in a volume at a temperature of 56 °C. Describe
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A flask at room temperature contains equal numbers of di-nitrogen molecules and krypton atoms. (a) Which of the two gases exerts the higher partial pressure? (b) Which gas has a higher kinetic energy per molecule/atom? (c) Which gas has molecules with a higher velocity? Explain your answers.
There are two particles, one is heavy and the other is light. The light particles diffuse faster than the heavy particles. This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is the velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas? If the light gas was Ne, what would be a reasonable identity for the heavy gas?
Chapter 5 Solutions
Chemistry
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